CharacteristicPolynomial - California State University...

Characteristic PolynomialMassoud Malek

♣ Preliminary Results. In all that follows, we denote the n × n identity matrix by In and then× n zero matrix by Zn.

Let A = (aij) be an n×n matrix. If Au = λu, then λ and u are called the eigenvalue and eigenvectorof A, respectively. The eigenvalues of A are the roots of the characteristic polynomial

KA(λ) = det (λ In −A ) .

The eigenvectors are the solutions to the Homogeneous system (λ In −A)X = θ.

Note that KA(λ) is a monic polynomial (i.e., the leading coefficient is one).

Cayley-Hamilton Theorem. If KA(λ) = λn + p1 λn−1 + · · ·+ pn−1 λ+ pn is the characteristic

polynomial of the n× n matrix A, then

KA (A) = An + p1An−1 + · · ·+ pn−1A+ pnIn = Zn,

Corollary. Let KA(λ) = λn + p1λn−1 + · · · + pn−1λ + pn be the characteristic polynomial of the

n× n invertible matrix A. Then

A−1 =1

−pn[An−1 + p1A

n−2 + · · ·+ pn−2A+ pn−1In].

Proof. According to the Cayley Hamilton’s theorem we have

A[An−1 + p1A

n−2 + · · ·+ pn−1 In]

= −pn In .

Since A is nonsingular, pn = (−1)n det(A) 6= 0; thus the result follows.

Newton’s Identity. Let λ1, λ2, . . . , λn be the roots of the polynomial

K(λ) = λn + p1λn−1 + p2λ

n−2 + · · · · · ·+ pn−1λ+ pn.

If sk = λk1 + λk2 + · · ·+ λkn, then

pk = −1

k(sk + sk−1 p1 + sk−2 p2 + · · ·+ s2 pk−2p1 + s1 pk−1) .

Proof. From K(λ) = (λ − λ1)(λ − λ2) . . . . . . (λ − λn−1)(λ − λn) and the use of logarithmicdifferentiation, we obtain

K ′(λ)

K (λ)=nλn−1 + (n− 1) c1λ

n−2 + · · ·+ 2 cn−2λ+ cn−1λn + c1λn−1 + c2λn−2 + · · ·+ cn−1λ+ cn

=

n∑i=1

1

(λ− λi).

By using the geometric series for1

(λ− λi)and choosing |λ| > max

1≤i≤n|λi|, we obtain

n∑i=1

1

(λ− λi)=n

λ+s1λ2

+s2λ3

+ · · · · · ·

Hence nλn−1 + (n− 1) p1 λn−2 + · · ·+ pn−1 =

(λn + p1 λ

n−1 + p2 λn−2 + · · ·+ pn

)(nλ

+s1λ2

+s2λ3

+ · · ·).

By equating both sides of the above equality we may obtain the Newton’s identities.

The Method of Direct Expansion Characteristic Polynomial 2

♣ The Method of Direct Expansion. The characteristic polynomial of an n× n matrix A =(aij) is defined as:

KA(λ)=det(λIn −A)=

∣∣∣∣∣∣∣∣∣λ− a11 −a12 . . . −a1n−a21 λ− a22 . . . −a2n

......

. . ....

−an1 −an2 . . . λ− ann

∣∣∣∣∣∣∣∣∣=λn−σ1λn−1+σ2λn−2−· · ·+ (−1)nσn,

where

σ1 =n∑

i=1

aii = trace(A)

is the sum of all first-order diagonal minors of A,

σ2 =∑i<j

∣∣∣∣aii aijaji ajj

∣∣∣∣is the sum of all second-order diagonal minors of A,

σ3 =∑

i<j<k

∣∣∣∣∣∣aii aij aikaji ajj ajkaki akj akk

∣∣∣∣∣∣is the sum of all third-order diagonal minors of A, and so forth. Finally,

σn = det(A)

There are

(n

k

)diagonal minors of order k in A. From this we find that the direct computation of

the coefficients of the characteristic polynomial of an n× n matrix is equivalent to computing(n

1

)+

(n

2

)+ · · ·

(n

n

)= 2n − 1

determinants of various orders, which, generally speaking, is a major task. This has given rise tospecial methods for expanding characteristic polynomial. We shall explain some of these methods.

Example. Compute the characteristic polynomial of A =

1 2 32 1 −41 0 2

.

We have:

σ1 = 1 + 1 + 2 = 4,

σ2 =

∣∣∣∣1 22 1

∣∣∣∣+

∣∣∣∣1 −40 2

∣∣∣∣+

∣∣∣∣1 31 2

∣∣∣∣ = (−3) + (2) + (−1) = −2,

σ3 = det(A) =

∣∣∣∣∣∣1 2 32 1 −41 0 2

∣∣∣∣∣∣ = −17.

ThusKA(λ) = det(λI3 −A) = λ3 − σ1λ2 + σ2λ− σ3 = λ3 − 4λ− 2λ+ 17.

Leverrier’s Algorithm Characteristic Polynomial 3

♣ Leverrier’s Algorithm. This method allows us to find the characteristic polynomial of any

n× n matrix A using the trace of the matrix Ak, where k = 1, 2, · · ·n. Let

σ(A) = {λ1, λ2, · · · , λn}

be the set of all eigenvalues of A which is also called the spectrum of A. Note that

sk = trace(Ak) =n∑

i=1

λki , for all k = 1, 2, · · · , n.

LetKA(λ) = det(λIn −A) = λn + p1λ

n−1 + · · ·+ pn−1λ+ pn

be the characteristic polynomial of the matrix A, then for k ≤ n, the Newton’s identities hold true:

pk = −1

k[sk + p1sk−1 + · · ·+ pk−1s1] (k = 1, 2, · · · , n)

Example. Let A =

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

. Then

A2 =

1 8 4 09 −1 −1 913 −12 5 815 −12 −6 7

A3 =

17 6 13 1942 −28 8 2343 −9 −16 2219 −11 −3 −17

A4 =

125 −48 16 104122 −23 −22 4690 −40 41 −12−66 120 0 −107

.So s1 = 4 , s2 = 12 , s3 = −44 , and s4 = 36. Hence

p1 = −s1 = −4,

p2 = −1

2(s2 + p1s1) = −1

2(12 + (−4)4) = 2,

p3 = −1

3(s3 + p1s2 + p2s1) = −1

3(−44 + (−4)12 + 2(4)) = 28,

p4 = −1

4(s4 + p1s3 + p2s2 + p3s1) = −1

4(36 + (−4)(−44) + 2(12) + 28(4)) = −87.

ThereforeKA(λ) = λ4 − 4λ3 + 2λ2 + 28λ− 87

and

A−1 =1

87

[A3 − 4A2 + 2A+ 28 I4

]=

1

87

17 6 13 1942 −28 8 2343 −9 −16 2219 −11 −3 −17

−4

1 8 4 09 −1 −1 913 −12 5 815 −12 −6 7

+2

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

+28

1 0 0 00 1 0 00 0 1 00 0 0 1

=1

87

43 −22 −1 178 4 16 −11−5 41 −10 −4−33 27 21 −9

.

The Method of Souriau Characteristic Polynomial 4

♣ The Method of Souriau (or Fadeev and Frame). This is an elegant modification ofthe Leverrier’s method.

Let A be an n× n matrix, then define

A1 = A, q 1 = −trace(A1), B1 = A1 + q1In;A2 = AB1, q 2 = −1

2 trace(A2), B2 = A2 + q2In;...

......

......

......

......

...An−1 = ABn−2, qn−1 = − 1

n−1 trace(An−1), Bn−1 = An−1 + qn−1In;

An = ABn−1, qn = − 1n trace(An), Bn = An + qnIn

.

Theorem. Bn = Zn, and

KA(λ) = det(λIn −A) = λn + q 1λn−1 + · · ·+ qn−1λ+ qn ,

where Adj (A) = Bn−1 and if A is nonsingular, then

A−1 = − 1

qnBn−1 .

Proof. Suppose the characteristic polynomial of A is

KA(λ) = det(λIn −A) = λn + p1λn−1 + · · ·+ pn−1λ+ pn ,

where p′ks are defined in the Leverrier’s method.

Clearly p1 = −trace(A) = −trace(A1) = q1, and now suppose that we have proved that

q1 = p1, q2 = p2, . . . , qk−1 = pk−1 .

Then by the hypothesis we have

Ak = ABk−1 = A(Ak−1 + qk−1In) = AAk−1 + qk−1A

= A[A(Ak−2 + qk−2In)] + qk−1A

= A2Ak−1 + qk−2A2 + qk−1A

=...

......

......

...

= Ak + q1Ak−1 + · · · + qk−1A .

Let si = trace(Ai) (i = 1, 2, . . . , k), then by Newton’s identities

−kqk = trace(Ak) = trace(Ak) + q1 trace(Ak−1) + · · ·+ qk−1 trace(A)

= sk + q1sk−1 + · · ·+ qk−1s1

= sk + p1sk−1 + · · ·+ pk−1s1

= −kpk .

showing that pk = qk. Hence this relation holds for all k.

By the Cayley-Hamilton theorem,

Bn = An + q1An−1 + · · ·+ qn−1A+ qnIn = Zn .

and soBn = An + qnIn = Zn; An = ABn−1 = −qnIn.

If A is nonsingular, then det(A) = (−1)nKA(0) = (−1)nqn 6= 0, and thus

A−1 = − 1

qnBn−1 .

The Method of Souriau Characteristic Polynomial 5

Example. Find the characteristic polynomial and if possible the inverse of the matrix

A =

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

.

For k = 1, 2, 3, 4, compute

Ak = ABk−1 qk =−1

ktrace(Ak), Bk = Ak + qkI4 .

A1 =

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

, q1 = −4, B1 =

−3 2 1 −11 −4 2 12 1 −5 34 −5 0 0

;

A2 =

−3 0 0 45 −1 −9 55 −16 9 −4−1 8 −6 −9

, q2 = 2, B2 =

−1 0 0 45 1 −9 55 −16 11 −4−1 8 −6 −7

;

A3 =

15 −22 −1 178 −24 16 −11−5 41 −38 −4−33 27 21 −37

, q3 = 28, B3 =

43 −22 −1 178 4 16 −11−5 41 −10 −4−33 27 21 −9

;

A4 =

87 0 0 00 87 0 00 0 87 00 0 0 87

, q4 = −87, B4 =

0 0 0 00 0 0 00 0 0 00 0 0 0

.

Therefore the characteristic polynomial of A is:

KA(λ) = λ4 − 4λ3 + 2λ2 + 28λ− 87.

Note that A4 is a diagonal matrix, so we only need to multiply the first row of A by the first columnof B3 to obtain 87. Since q4 = −87, the matrix A has an inverse.

A−1 =−1

q4B3 =

1

87

43 −22 −1 178 4 16 −11−5 41 −10 −4−33 27 21 −9

.

Matlab ProgramA = input(′Enter a square matrix : ′)

m = size(A); n = m(1); q = zeros(1, n); B = A; AB = A; In = eye(n);

for k = 1 : n− 1, q(k) = −(1/k) ∗ trace(AB)B = AB + q(k) ∗ In; AB = A ∗B; end

C = B; q(n) = −(1/n) ∗ trace(AB); Q = [1 q];

disp(′The Characteristic polynomial looks like : ′)

disp( ′KA(x) = x ∧ n+ q(1)x ∧ (n− 1) + ...+ q(n− 1)x+ q(n)′ ), disp(′ ′),

disp(′The coefficients list c(k) is : ′), disp(′ ′),

disp(Q), disp(′ ′)

if q(n) == 0, disp(′The matrix is singular ′);

else, disp(′The matrix has an inverse. ′), disp(′ ′)

C = −(1/q(n)) ∗B;

disp(′The inverse of A is : ′), disp(′ ′),

disp(C)

end

The Method of Undetermined Coefficients Characteristic Polynomial 6

♣ The Method of Undetermined Coefficients. If one has to expand large numbers of char-acteristic polynomials of matrices of the same order, then the method of undetermined coefficientsmay be used to produce characteristic polynomials of those matrices.

Let A be an n× n matrix and

KA(λ) = det (λ In −A) = λn + p1λn−1 + · · ·+ pn−1λ+ pn .

be its characteristic polynomial. In order to find the coefficients pi’s of KA(λ) we evaluate

Dj = KA(j) = det (jIn −A) j = 0, 1, 2, . . . , n− 1

and obtain the following system of linear equations:

pn = D0

1n + p1.1n−1 + · · · · · · · · · · · · · · ·+ pn = D1

2n + p1.2n−1 + · · · · · · · · · · · · · · ·+ pn = D2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(n− 1)n + p1.(n− 1)n−1 + · · ·+ pn = Dn−1

Which can be changed into:

Sn−1P =

1n−1 1n−2 . . . 12n−1 2n−2 . . . 2

......

. . ....

(n− 1)n−1 (n− 1)n−2 . . . n− 1

p1p2...

pn−1

=

D1 −D0 − 1n

D2 −D0 − 2n

......

......

Dn−1 −D0 − (n− 1)n

= D .

The system may be solved as follows:P = S −1n D .

Since the (n− 1)× (n− 1) matrix Sn depends only on the order of A, we may store Rn, the inverseof Sn−1 beforehand and use it to find the coefficients of characteristic polynomial of various n × nmatrices.

Examples. Compute the characteristic polynomials of the 4× 4 matrices

A =

1 3 0 42 −3 1 31 2 1 2−1 3 2 1

and B =

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

.

First we find

S =

1 1 18 4 227 9 3

and R = S−1 = − 1

12

−6 6 −230 −24 6−36 18 −4

.

Then for the matrix A we obtain

D0 = det(−A) = −48, D1 = det(I4 −A) = −72,

D2 = det(2I4 −A) = −128, and D3 = det(3I4 −A) = −180 .

D =

D1 −D0 − 14

D2 −D0 − 24

D3 −D0 − 34

=

−25−96−213

.

Hence

P =

p1p2p3

= − 1

12

−6 6 −230 −24 6−36 18 −4

−25−96−213

=

0−23−2

.

The Method of Undetermined Coefficients Characteristic Polynomial 7

ThusKA(λ) = λ4 − 23λ2 − 2λ− 48.

For the matrix B we have

D0 = det(−B) = −87, D1 = det(I4 −B) = −60,

D2 = det(2I4 −B) = −39, and D3 = det(3I4 −B) = −12.

D =

D1 −D0 − 14

D2 −D0 − 24

D3 −D0 − 34

=

2632−6

.

Hence

P =

p1p2p3

= − 1

12

−6 6 −230 −24 6−36 18 −4

2632−6

=

−4228

.

ThusKB(λ) = λ4 − 4λ3 + 2λ2 + 28λ− 87.

Matlab Program

N = input(‘Enter the size of your square matrix : ′);

n = N − 1; In = eye(N); S = zeros(n); R = zeros(n); D = zeros(1, n);

DSP1 = [ ′ For any ′, int2str(N),′−square matrix, you need S =′ ];

DSP2 = [ ′Do you want to try with another ′, int2str(N),′−square matrix? (Y es = 1/No = 0)′ ];

%DEFINING S

for i = 1 : n, for j = 1 : n, S(i, j) = i ∧ (N − j); end; end;

disp(′ ′), disp(DSP1), disp(′ ′), disp(S),

R = inv(S);

ok = 1;

while ok == 1;

A = input([′Enteran ′, int2str(N),′ x ′, int2str(N),′ matrix A : ′); disp(′ ′)

D0 = det(A);

for k = 1 : n; D(k) = det(k ∗ In−A); end;

for i = 1 : n; DD(i) = D(i)−D0− i ∧N ; end;

P = R ∗DD′;disp(′The Characteristic polynomial looks like : ′)

disp( ′KA(x) = x ∧ n+ p(1)x ∧ (n− 1) + ...+ p(n− 1)x+ p(n)′ ), disp(′ ′),

disp(′The coefficients list p(k) is : ′), disp(′ ′),

disp([1 P ′ D0]), disp(′ ′),

disp(DSP2), disp(′ ′),

ok = input(DSP2);

end

The Method of Danilevsky Characteristic Polynomial 8

♣ The Method of Danilevsky. Consider an n× n matrix A and let

KA(λ) = det(λIn −A) = λn + p1λn−1 + · · ·+ pn−1λ+ pn

be its characteristic polynomial. Then the companion matrix of KA(λ)

F [A] =

−p1 −p2 −p3 . . . −pn−1 −pn1 0 0 . . . 0 00 1 0 . . . 0 0...

.... . .

. . ....

...0 0 . . . 1 0 00 0 0 . . . 1 0

is similar to A and is called the Frobenius form of A.

The method of Danilevsky (1937) applies the Gauss-Jordan method to obtain the Frobenius formof an n × n matrix. According to this method the transition from the matrix A to F [A] is doneby means of n− 1 similarity transformations which successively transform the rows of A, beginningwith the last, into corresponding rows of F [A].

Let us illustrate the beginning of the process. Our purpose is to carry the nth row of

A =

a11 a12 a13 . . . a1,n−1 a1na21 a22 a23 . . . a2,n−1 a2na31 a32 a33 . . . a3,n−1 a3n...

......

. . ....

...an1 an2 an3 . . . an,n−1 ann

into the row

(0 0 . . . 1 0

). Assuming that an,n−1 6= 0, we replace the (n−1)th row of the n×n

identity matrix with the nth row of A and obtained the matrix

Un−1 =

1 0 0 . . . 0 00 1 0 . . . 0 0...

.... . . . . .

......

an1 an2 an3 . . . an,n−1 ann0 0 0 . . . 0 1

.

The inverse of Un−1 is

Vn−1 =

1 0 0 . . . 0 00 1 0 . . . 0 0...

.... . . . . .

......

vn−1,1 vn−1,2 vn−1,3 . . . vn−1,n−1 vn−1,n0 0 0 . . . 0 1

.

wherevn−1,i = − ani

an,n−1for i 6= n− 1

and

vn−1,n−1 =1

an,n−1.

Multiplying the right side of A by Vn−1, we obtain

AVn−1 = B =

b11 b12 b13 . . . b1,n−1 b1nb21 b22 b23 . . . b2,n−1 b2n...

.... . . . . .

......

bn−1,1 bn−1,2 bn−1,3 . . . bn−1,n−1 bn−1,n−10 0 0 . . . 1 0


However the matrix B = AMn−1 is not similar to A. To have a similarity transformation, it isnecessary to multiply the left side of B by Un−1 = V −1n−1. Let C = Un−1AVn−1, then C is similar toA and is of the form

C =

c11 c12 c13 . . . c1,n−1 c1nb21 b22 b23 . . . b2,n−1 b2n...

.... . . . . .

......

cn−1,1 cn−1,2 cn−1,3 . . . cn−1,n−1 cn−1,n−10 0 0 . . . 1 0

Now, if cn−1,n−1 6= 0, then similar operations are performed on matrix C by taking its (n− 2)th rowas the principal one. We then obtain the matrix

D = Un−2CVn−2 = Un−2Un−1AVn−1Vn−2

with two reduced rows. We continue the same way until we finally obtain the Frobenius form

F [A] = U1U2 · · ·Un−2Un−1AVn−1Vn−2 · · ·V2V1 .

if, of course, all the n− 1 intermediate transformations are possible.

Exceptional case in the Danilevsky method. Suppose that in the transformation of the matrixA into its Frobenius form F [A] we arrived, after a few steps, at a matrix of the form

R =

r11 r12 . . . r1k . . . r1,n−1 r1nr21 r22 . . . r2k . . . r2,n−1 r2n...

......

......

......

rk1 rk2 . . . rkk . . . rk,n−1 rkn0 0 . . . 1 . . . 0 00 0 . . . 0 . . . 0 0...

......

......

......

0 0 . . . 0 . . . 1 0

and it was found that rk,k−1 = 0 or |rk,k−1| is very small. It is then possible to continue thetransformation by the Danilevsky method.

Two cases are possible here.

Case 1. Suppose for some j = 1, 2, . . . , k−2, rkj 6= 0. Then by permuting the jth row and (k−1)throw and the jth column and (k− 1)th column of R we obtain a matrix R′ = (r′ij) similar to R withr′k,k−1 6= 0.

Case 2. Suppose now that rkj = 0 for all j = 1, 2, . . . , k − 2. Then R is in the form

R =

[R1 R2

0 R3

]=

r11 r12 . . . r1,k−1 r1,k r1,k+1 . . . r1nr21 r22 . . . r2,k−1 r2,k r2,k+1 . . . r2n...

.... . .

......

. . ....

...rk−1,1 rk−1,2 . . . rk−1,k−1 rk−1,k rk−1,k+1 . . . rk−1,n

......

. . ....

.... . .

......

0 0 . . . 0 rkk rk,k+1 . . . rkn0 0 . . . 0 1 . . . 0 0...

.... . .

......

. . ....

...0 0 . . . 0 0 . . . 1 0

.

In this case the characteristic polynomial of R breaks up into two determinants:

det(λIn −R) = det(λIk−1 −R1) det(λIn−k+1 −R3).


Here, the matrix R3 is already reduced to the Frobenius form. It remains to apply the Danilevsky’smethod to the matrix R1.

Note. Since Uk Ak−1 only changes the kth row of Ak−1, it is more efficient to multiply first Ak−1by its (k + 1)th row and then multiply on the right side the resulting matrix by Vk.

The next result shows that once we transform A into its Frobenius form ; we may obtain theeigenvectors with the help of the matrices V ′i s.

Theorem. Let A be an n × n matrix and let F [A] be its Frobenius form. If λ is an eigenvalue ofA, then

v =

λn−1

λn−2

...λ1

and w = Vn−1Vn−2 · · ·V2V1v

are the eigenvectors of F [A] and A respectively.

Proof. Since

det(λIn −A) = det(λIn − F [A] ) = λn + p1λn−1 + · · ·+ pn−1λ+ pn ,

we have

(λIn − F [A])v =

λ− p1 −p2 −p3 . . . −pn−1 −pn

1 λ 0 . . . 0 0λ 1 0 . . . 0 0...

......

. . ....

...0 0 0 . . . 1 λ

λn−1

λn−2

...λ1

=

000...0

.

Since F [A] = V −11 V −12 · · ·V −1n−2V−1n−1AVn−1Vn−2 · · ·V2V1 and F [A]v = λv, we conclude that

λw = Vn−2 · · ·V2 V1(λv) = (Vn−2 · · ·V2 V1) F [A]v = A (Vn−1Vn−2 · · ·V2 V1v) = Aw

Note. For expanding characteristic polynomials of matrices of order higher than fifth, the methodof Danilevsky requires less multiplications and additions than other methods.

Example. Reduce the matrix

A =

1 1 3 42 0 2 11 0 1 20 0 −1 −1

to its Frobenius form.

The matrix B = A3 = U3AV3 is as follows:

B =

1 0 0 00 1 0 00 0 −1 −10 0 0 1

1 1 3 42 0 2 11 0 1 20 0 −1 −1

1 0 0 00 1 0 00 0 −1 −10 0 0 1

=

1 1 −3 12 0 −2 −1−1 0 0 −10 0 1 0

.

Since b32 = 0, we need the permutation matrix J =

0 1 0 01 0 0 00 0 1 00 0 0 1

; thus

C = J B J =

0 1 0 01 0 0 00 0 1 00 0 0 1

1 1 −3 12 0 −2 −11 0 −1 10 0 1 0

0 1 0 01 0 0 00 0 1 00 0 0 1

=

0 2 −2 −11 1 −3 10 −1 0 −10 0 1 0

.


Next we obtain the matrix D = A2 = U2C V2

D =

1 0 0 00 −1 0 −10 0 1 00 0 0 1

0 2 −2 −11 1 −3 10 −1 0 −10 0 1 0

1 0 0 00 −1 0 −10 0 1 00 0 0 1

=

0 −2 −2 −3−1 1 2 00 1 0 00 0 1 0

.

Finally the Frobenius form F [A] = A1 = U1DV1,

F [A] =

−1 1 2 00 1 0 00 0 1 00 0 0 1

0 −2 −2 −3−1 1 2 00 1 0 00 0 1 0

−1 1 2 00 1 0 00 0 1 00 0 0 1

=

1 4 2 31 0 0 00 1 0 00 0 1 0

.

Thus the Characteristic polynomial of A is KA(λ) = x4 − x3 − 4x2 − 2x− 3

Using MATLAB for the Danilevsky Method

Consider the matrix A:

>> A = [ 1 2 4 3 ; 2 4 5 1 ; 3 2 1 4 ; 5 1 2 3 ] , M = A ; I = eye(4);

A =

1 2 4 32 4 5 13 2 1 45 1 2 3

Use the fourth row of A to define U and its inverse V:

>> U = I ; U (3, :) = A (4, :) , V = I ; V (3, :) = −A (4, :) /A (4, 3) ; V (3, 3) = 1/A (4, 3)

U =

1 0 0 00 1 0 05 1 2 30 0 0 1

V =

1.0000 0.0000 0.0000 0.00000.0000 1.0000 0.0000 0.0000−2.5000 −0.5000 0.5000 −1.50000.0000 0.0000 0.0000 1.0000

Define the matrix B similar to A which has the same characteristic polynomial.

>> B = U ∗ A ∗ V

B =

−9.0000 0.0000 2.0000 −3.0000−10.5000 1.5000 2.5000 −6.5000−54.5000 4.5000 16.5000 −16.5000

0.0000 0.0000 1.0000 0.0000

Change B into A and find a new U and its inverse V, by using the third row of the new A:

>> A = B ; U = I ; U (2, :) = A (3, :) , V = I ; V (2, :) = −A (3, :) /A (3, 2) ;V (2, 2) = 1/A (3, 2)

U =

1.0000 0.0000 0.0000 0.0000−54.5000 4.5000 16.5000 −16.5000

0.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000

V =

1.0000 0.0000 0.0000 0.000012.1111 0.2222 −3.6667 3.66670.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000



>> B = U ∗ A ∗ V

B =

−9 0 2 −3525 18 −139 1590 1 0 00 0 1 0

Change B into A and find a new U and its inverse V, using the second row of the new A:

>> A = B ; U = I ; U (1, :) = A (2, :) , V (1, :) = −A (2, :) /A (2, 1) ;V (1, 1) = 1/A(2, 1)

U =

−10.5000 1.5000 2.5000 −6.5000

0.0000 1.0000 0.0000 0.00000.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000

V =

0.0019 −0.0343 0.2648 −0.30290.0000 1.0000 0.0000 0.00000.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000

>> B = U ∗ A ∗ V

B =

9 23 −42 −1441 0 0 00 1 0 00 0 1 0

B is the companion matrix of our original matrix A. Here is the characteristic polynomial of A:

1.0000 − 9.0000 − 23.0000 42.0000 144.0000

Exceptional Case.

>> A = [ 1 2 4 3 ; 2 4 5 1 ; 3 2 1 4 ; 5 1 0 3 ]

A =

1 2 4 32 4 5 13 2 1 45 1 0 3

Since A (4, 3) = 0 , we need a permutation matrix P which moves A (4, 3) = 0 into another position:

>> P = I ; Q = P P (4, :) = P (3, :) P (3, :) = Q (4, :)

P =

1 0 0 00 1 0 00 0 0 10 0 1 0

Define the matrix B similar to A which has the same characteristic polynomial, but B (4, 3) 6= 0.

Note that every time there is a zero in A (k, k+ 1) entry, use a permutation matrix, to obtain a newmatrix similar the matrix A, but with a non-zero value at that position.

>> B = P ∗ A ∗ P ′

B =

1 2 3 42 4 1 55 1 3 03 2 4 1

We set A to B and continue the same way as the previous example.


>> U = I ; U (3, :) = A (4, :) , V = I ; V (3, :) = −A (4, :) /A (4, 3) ; V (3, 3) = 1/A (4, 3)

U =

1 0 0 00 1 0 03 2 4 10 0 0 1

V =

1.0000 0.0000 0.0000 0.00000.0000 1.0000 0.0000 0.0000−0.7500 −0.5000 0.2500 −0.25001.0000 0.0000 0.0000 1.0000


>> B = U ∗ A ∗ V

B =

−1.2500 0.5000 0.7500 3.25001.2500 3.5000 0.2500 4.75009.7500 6.5000 6.7500 16.25000.0000 0.0000 1.0000 0.0000


>> A = B ; U = I ; U (2, :) = A (3, :) , V = I ; V (2, :) = −A (3, :) /A (3, 2) ; V (2, 2) = 1/A (3, 2)

U =

1.0000 0.0000 0.0000 0.0000−45.5000 11.0000 −3.5000 −6.5000

0.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000

V =

1.0000 0.0000 0.0000 0.00004.1364 0.0909 0.3182 0.59090.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000


>> B = U ∗ A ∗ V

B =

−1.6818 0.0070 0.2552 2.045557.0227 −3.2797 −53.4598 −133.56820.0000 1.0000 0 0

0 0 1.0000 0

Note that A (2, 1) 6= 0, so we don’t need any permutation matrix.


>> A = B ; U = I ; U (1, :) = A (2, :) , V = I ; V (1, :) = −A (2, :) /A (2, 1) ; V (1, 1) = 1/A (2, 1)

U =

57.0227 −3.2797 −53.4598 −133.56820.0000 1.0000 0.0000 0.00000.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000

V =

0.0175 0.0575 0.9375 2.34240.0000 1.0000 0.0000 0.00000.0000 0.0000 1.0000 0.00000.0000 0.0000 0.0000 1.0000


>> B = U ∗ A ∗ V

B =

−4.9615 −58.5769 −208.9231 −108.00001.0000 0.0000 0.0000 0.00000.0000 1.0000 0.0000 0.00000.0000 0.0000 1.0000 0.0000

B is the companion matrix of our original matrix A. Here is the characteristic polynomial of A:

1.0000 4.9615 58.5769 208.9231 108.0000


Matlab Program

A = input(′Enter the square matrix A : ′);

m = size(A); N = m(1); b = [1]; B = zeros(N); i = 1;

while i < N,

J = eye(N);h = A(N − i+ 1, N − i)while h == 0;

c = A(N − i+ 1, 1 : N − i); z = norm(c, inf);

if z = 0;

k = 1; r = 0;

while r == 0 & k < N − i;r = r + c(N − i− k); k = k + 1;

J(N − i,N − i) = 0; J(N − i,N − i− k + 1) = 1;

J(N − i− k + 1, N − i− k + 1) = 0; J(N − i− k + 1, N − i) = 1,

A = J ∗A ∗ J ; k = k + 1;

endelse

b = conv(b, [1−A(N − i+ 1, N − i+ 1 : N)]);

B = A(1 : N − i, 1 : N − i);A = B,N = N − i;end

h = A(N − i+ 1, N − i);end

U = eye(N); V = eye(N);

U(N − i, :) = A(N − i+ 1, :);

V (N − i, :) = −A(N − i+ 1, :)/A(N − i+ 1, N − i); V (N − i,N − i) = 1/A(N − i+ 1, N − i);A = U ∗A ∗ V ;

i = i+ 1;

end;

b = conv(b, [1−A(N − i+ 1, N − i+ 1 : N)]); disp(′ ′),

disp(′The Characteristic polynomial looks like : ′), disp(′ ′),

disp([′KA(x) = x ∧ n+ c(1)x ∧ (n− 1),+...+ c(n− 1)x+ c(n)′]), disp(′ ′),

disp(′The coefficients list c(k) is : ′), disp(′ ′),

disp(b), disp(′ ′)

The Method of Krylov Characteristic Polynomial 15

♣ The Method of Krylov. Let A be an n× n matrix. For any n-dimensional nonzero columnvector v we associate its successive transforms

vk = Akv (k = 0, 1, 2, . . .);

this sequence of vectors is called the Krylov sequence associated to the matrix A and the vector v.

At most n vectors of the sequence v0, v1, v2, . . . will be linearly independent.

Suppose for some r = r(v) ≤ n, the vectors v0 , v1 , v2 , . . . , vr are linearly independent and thevector vr+1 is a linear combination of the preceding ones. Hence there exists a monic polynomial

φ(λ) = c0 + c1λ+ c2λ2 + · · ·+ cr−1λ

r−1 + λr

such that

φ(A) v = (c0 In + c1A+ · · ·+ cr−1Ar−1 +Ar) v = c0 v0 + c1 v1 + · · ·+ cr−1 vr + cr vr+1 = θ .

The polynomial φ(λ) is said to annihilate v and to be minimal for v. If ω (λ) is another monicpolynomial which annihilates v,

ω(A) v = θ ,

then φ (λ) divides ω (λ).

To show that; supposeω (λ) = γ (λ)φ (λ) + ρ (λ),

where ρ (λ) is the remainder after dividing ω by φ, hence of degree strictly less than ρ, it followsthat

ρ(A) v = θ .

But φ (λ) is minimal for v, hence ρ (λ) = 0.

Now of all vectors v there is at least one vector for which the degree v is maximal, since for anyvector v, r(v) ≤ n. We call such vector a maximal vector.

Remark. A monic polynomial µA (λ) of an n× n matrix A is said to be its minimal polynomial, ifµA (λ) is of minimum degree satisfying

µ (A) = Zn.

If the minimal polynomial of a matrix is equal to its characteristic polynomial, then we may useKrylov method to find the characteristic polynomial. For example, if the matrix has n distincteigenvalues, then its minimal polynomial is equal to its characteristic polynomial. Hence the Krylovmethod would be successful.

Algorithm. To produce the characteristic polynomial of the matrix A by Krylov method, wefollow the following steps:

Step 1. Choose an arbitrary n-dimensional nonzero column vector v such as e1, then use the Krylovsequence to define the matrix

V =[v , A v , A2 v , . . . , An−2 v , An−1 v

]= [ v0 , v1 , v − 2 , . . . vn−2 , vn−1] .

Step 2. If the matrix V has rank n, then the system V c = −vn has a unique solution

ct = (c0 , c1 , c2 , · · · , cn−1).

The monic polynomial

φ(λ) = c0 + c1 λ+ c2 λ2 + · · ·+ cn−1 λ

n−1 + λn

which annihilates the vector v is the characteristic polynomial of A.

Step 3. If the system V c = −vn does not have a unique solution, then change the initial vector


and try for example with e2.

Step 4. If again, the system V c = −vn does not have a unique solution, then either chooseanother initial vector v or suspect that the minimal polynomial and characteristic polynomial of Aare different, in this case abandon this method and use another method.

Examples. Compute the characteristic polynomials of the following matrices:

A =

1 2 1 −11 0 2 12 1 −1 34 −5 0 4

, B =

1 2 −3 11 0 −2 11 −3 −1 31 0 1 −2

, and C =

1 2 3 41 2 3 41 0 0 01 0 0 0

.

Choosing the initial vector v =

1000

for the matrices A and B, we obtain

VA =[v , A v , A2 v , A3 v

]=

1 1 1 170 1 9 420 2 13 430 4 15 19

and VB =[v , B v , B2 v , B3 v

]=

1 1 1 10 1 0 10 1 0 10 1 0 1

.

The matrix VA is nonsingular, hence cA = −V −1A A4v =

−87282−4

. From the vector cA we obtain the

characteristic polynomial of A which is

KA (λ) =(1 , λ , λ2 , λ4

)·

−87282−4

+ λ4 = −87 + 28λ+ 2λ2 − 4λ3 + λ4.

The matrix VB is singular, so we need another initial vector such as v =

0100

. The new matrix

VB =

0 2 11 111 0 8 00 −3 5 −210 0 −1 18

is invertible, so cB = −V −1B B4 v =

9−2−10

2

. From the vector cB we

obtain the characteristic polynomial of B which is

KB(λ) =(1 , λ , λ2 , λ4

)·

9−2−10

2

+ λ4 = 9− 2λ− 10λ2 + 2λ3 + λ4.

The minimal and characteristic polynomials of the matrix C are

mC (λ) = λ3 − 3λ2 − 7λ and KC (λ) = λ4 − 3λ3 − 7λ2,

respectively. Therefore by choosing any initial vector v, the matrix VC =[v , C v , C2 v , C3 v

]will always be singular. This means that the Krylov sequence will never produce the characteristicpolynomial KC (λ).


Matlab Program

A = input(′Enter a square matrix A : ′);

m = size(A); n = m(1); V = zeros(n, n);

DL1 = [′Enter an initial ′, int2str(n) , ′ − dimensional row vector v0 = ′];

v0 = input(DL1);

z = 0; k = 1;

while z == 0 & k < 5

w = v0; V (:, 1) = w;

for i = 2 : n, w = A ∗ w; V (:, i) = w; end,

if det(V ) ∼= 0; k = 8; c = −inv(V ) ∗A ∗ w;

else

while k < 5

v0 = input(′The matrix V is singular, please enter another initial row vector v0 : ′);

k = k + 1;

end;

end;

z = det(V );

end;

if k == 5;

disp(′Sorry, the Krylov method is not suited for this matrix. ′), disp(′ ′),

else;

disp(′The Characteristic polynomial looks like : ′), disp(′ ′),

disp([′KA(x) = c(0) + C(1)x+ c(2)x ∧ 2 + · · ·+ c(n− 1)x ∧ (n− 1) + x ∧ n′]), disp(′ ′),disp(′The coefficients list c(k) is : ′), disp(′ ′),

disp([c′, 1])

end;

CharacteristicPolynomial - California State University...

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